Simplify the following expression and state the condition under which the simplification is valid. You can assume that $a \neq 0$. $z = \dfrac{-3}{3(3a + 7)} \times \dfrac{6(3a + 7)}{7a} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ -3 \times 6(3a + 7) } { 3(3a + 7) \times 7a } $ $ z = \dfrac{-18(3a + 7)}{21a(3a + 7)} $ We can cancel the $3a + 7$ so long as $3a + 7 \neq 0$ Therefore $a \neq -\dfrac{7}{3}$ $z = \dfrac{-18 \cancel{(3a + 7})}{21a \cancel{(3a + 7)}} = -\dfrac{18}{21a} = -\dfrac{6}{7a} $